Show an element is not in a ring of fractions

Question

Let $(\mathbb{R}[x,y]/(x^2-y^3))_{(x,y)}$. How could I show that $\frac{x}{y}$ does not belong to the ring before?

I think I could prove it by using the definition of the ring of fractions, but is there a simple way to do it?

Answer 1

It's possible to do this "entirely by hand", which I think is a good way to do it at least once.

When we ask "does $x/y$ belong to the ring $(\mathbb{R}[x,y]/(x^2-y^3))_{(x,y)}$?", what we're really asking is "does there exist some element $a \in (\mathbb{R}[x,y]/(x^2-y^3))_{(x,y)}$ such that $ay = x$?". To answer that question, suppose such an $a$ did exist. Then by construction of the localization, $a = p/q$ for some $p,q \in \mathbb{R}[x,y]/(x^2-y^3)$ such that $q \notin (x,y)$. Then $ay = x$ means that $(yp)/q = x/1$, which means that there exists some $t \notin (x,y)$ such that $t(yp-xq) = 0$.

Of course this equality is happening in the quotient ring $\mathbb{R}[x,y]/(x^2-y^3)$, and it's easier to think about genuine polynomials, so let's take representatives $\hat{t}, \hat{p}, \hat{q}$ of $t, p, q$ (respectively) in $\mathbb{R}[x,y]$, and now our condition is that $\hat{t}(y \hat{p} - x \hat{q}) \in (x^2 - y^3)$. Thus there is some $b \in \mathbb{R}[x,y]$ such that $\hat{t}(y \hat{p} - x \hat{q}) = b(x^2 - y^3)$. This is now a question of polynomial arithmetic: does there exist a solution? Remember in the end we need $t, q \notin (x,y)/(x^2 - y^3)$.